Skip to main content

computational geometry - Smooth convex hull of a large data set of 3D points


I recently needed to deal with a large data set of 600,000 points in three dimensions. My task was to find a convex hull for this data. I have Mathematica 10, so I could use the function ConvexHullMesh; I obtained this:


screenshot convex hull



I was wondering if there is some way to find a smooth convex hull (maybe an ellipsoid) for my data.



Answer



Minimum Volume Ellipsoid


Translated from here, this uses the Khachiyan algorithm, and should work for any dimension.


    MinVolEllipse[P_, tolerance_] := 
     Module[{d, n, Q, count, err, u, X, M, maximum, j, stepSize, newu, U, A, c},

      {n, d} = Dimensions[P];
      Q = Append[1] /@ P;


      count = 1;
      err = 1;
      u = ConstantArray[1./n, n];

      While[err > tolerance,
       X = Q\[Transpose].DiagonalMatrix[u].Q;
       M = Diagonal[Q.Inverse[X].Q\[Transpose]];
       maximum = Max[M];
       j = Position[M, maximum][[1, 1]];
       stepSize = (maximum - d - 1)/((d + 1) (maximum - 1));

       newu = (1 - stepSize) u;
       newu[[j]] += stepSize;
       count += 1;
       err = Norm[newu - u];
       u = newu;
       ];

      U = DiagonalMatrix[u];

      A = (1/d) Inverse[P\[Transpose].U.P - Outer[Times, u.P, u.P]];

      c = u.P;

      {c, A}
      ]

Usage:


    pts = RandomVariate[
       MultinormalDistribution[RandomReal[{-1, 1}, {2}], 
        With[{m = RandomReal[{0, 1}, {2, 2}]}, m.m\[Transpose]]], 500];


    P = MeshCoordinates[ConvexHullMesh[pts]];
    tolerance = 0.0001;

    {c, A} = MinVolEllipse[P, tolerance];

    X = {x, y};
    Show[
     ConvexHullMesh[pts],
     Graphics[{
       Point[pts],

       {Red, Point[P]}
       }],
     ContourPlot[(X - c).A.(X - c) == 1, {x, -4, 4}, {y, -4, 4}]
     ]

enter image description here


In 3D:


    pts = RandomVariate[
       MultinormalDistribution[RandomReal[{-1, 1}, {3}], 
        With[{m = RandomReal[{0, 1}, {3, 3}]}, m.m\[Transpose]]], 100];


    P = MeshCoordinates[ConvexHullMesh[pts]];
    tolerance = 0.0001;

    {c, A} = MinVolEllipse[P, tolerance];

    X = {x, y, z};
    Show[
     ConvexHullMesh[pts, MeshCellStyle -> {{2, All} -> Opacity[0.5, Green]}],
     Graphics3D[{

       Point[pts],
       {Red, Point[P]}
       }],
     ContourPlot3D[(X - c).A.(X - c) == 1, {x, -3, 3}, {y, -3, 3}, {z, -3, 3},
      ContourStyle -> Directive[Red, Opacity[0.2]]
      ]
     ]

enter image description here


Comments

Post a Comment

Popular posts from this blog

functions - Get leading series expansion term?

Given a function f[x] , I would like to have a function leadingSeries that returns just the leading term in the series around x=0 . For example: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x)] x and leadingSeries[(1/x + 2 + (1 - 1/x^3)/4)/(4 + x)] -(1/(16 x^3)) Is there such a function in Mathematica? Or maybe one can implement it efficiently? EDIT I finally went with the following implementation, based on Carl Woll 's answer: lds[ex_,x_]:=( (ex/.x->(x+O[x]^2))/.SeriesData[U_,Z_,L_List,Mi_,Ma_,De_]:>SeriesData[U,Z,{L[[1]]},Mi,Mi+1,De]//Quiet//Normal) The advantage is, that this one also properly works with functions whose leading term is a constant: lds[Exp[x],x] 1 Answer Update 1 Updated to eliminate SeriesData and to not return additional terms Perhaps you could use: leadingSeries[expr_, x_] := Normal[expr /. x->(x+O[x]^2) /. a_List :> Take[a, 1]] Then for your examples: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x), x] leadingSeries[Exp[x], x] leadingSeries[(1/x + 2 + (1 - 1/x...
Post a Comment
Read more

mathematical optimization - Minimizing using indices, error: Part::pkspec1: The expression cannot be used as a part specification

I want to use Minimize where the variables to minimize are indices pointing into an array. Here a MWE that hopefully shows what my problem is. vars = u@# & /@ Range[3]; cons = Flatten@ { Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; Minimize[{Total@((vec1[[#]] - vec2[[u[#]]])^2 & /@ Range[1, 3]), cons}, vars, Integers] The error I get: Part::pkspec1: The expression u[1] cannot be used as a part specification. >> Answer Ok, it seems that one can get around Mathematica trying to evaluate vec2[[u[1]]] too early by using the function Indexed[vec2,u[1]] . The working MWE would then look like the following: vars = u@# & /@ Range[3]; cons = Flatten@{ Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; NMinimize[ {Total@((vec1[[#]] - Indexed[vec2, u[#]])^2 & /@ R...
Post a Comment
Read more

What is and isn't a valid variable specification for Manipulate?

I have an expression whose terms have arguments (representing subscripts), like this: myExpr = A[0] + V[1,T] I would like to put it inside a Manipulate to see its value as I move around the parameters. (The goal is eventually to plot it wrt one of the variables inside.) However, Mathematica complains when I set V[1,T] as a manipulated variable: Manipulate[Evaluate[myExpr], {A[0], 0, 1}, {V[1, T], 0, 1}] (*Manipulate::vsform: Manipulate argument {V[1,T],0,1} does not have the correct form for a variable specification. >> *) As a workaround, if I get rid of the symbol T inside the argument, it works fine: Manipulate[ Evaluate[myExpr /. T -> 15], {A[0], 0, 1}, {V[1, 15], 0, 1}] Why this behavior? Can anyone point me to the documentation that says what counts as a valid variable? And is there a way to get Manpiulate to accept an expression with a symbolic argument as a variable? Investigations I've done so far: I tried using variableQ from this answer , but it says V[1...
Post a Comment
Read more